Download 6.5 Dividing Polynomials

LESSON
6.5
Dividing Polynomials
Name
Class
6.5
HARDBOUND SE
PAGE 281
BEGINS HERE
Date
Dividing Polynomials
Essential Question: What are some ways to divide polynomials, and how do you know when
the divisor is a factor of the dividend?
Common Core Math Standards
The student is expected to:
A-APR.6
Evaluating a Polynomial Function
Using Synthetic Substitution
Explore
Rewrite simple rational expressions in different forms; write a(x)/b(x) in
the form q(x) + r(x)/b(x), where a(x), b(x), q(x), and r(x) are polynomials
with the degree of r(x) less than the degree of b(x), using inspection, long
division, or, for the more complicated examples, a computer algebra
system. Also A-APR.1, A-APR.3
Polynomials can be written in something called nested form. A polynomial in nested form is written in such a way
that evaluating it involves an alternating sequence of additions and multiplications. For instance, the nested form of
p(x) = 4x 3 + 3x 2 + 2x + 1 is p(x) = x(x(4x + 3) + 2) + 1, which you evaluate by starting with 4, multiplying by
the value of x, adding 3, multiplying by x, adding 2, multiplying by x, and adding 1.
Mathematical Practices
A
MP.2 Reasoning
Given p(x) = 4x 3 + 3x 2 + 2x + 1, find p(-2).
Rewrite p(x) as p(x) = x(x(4x + 3) + 2) + 1.
Language Objective
Multiply.
Work in small groups to complete a compare and contrast chart for
dividing polynomials.
Add.
Multiply.
-2 · 4 = -8
-8 + 3 = -5
-5 · -2 = 10
10 + 2 = 12
Add.
ENGAGE
Multiply.
Essential Question: What are some
ways to divide polynomials, and how do
you know when the divisor is a factor of
the dividend?
You can set up an array of numbers that captures the sequence of multiplications and
additions needed to find p(a). Using this array to find p(a) is called synthetic substitution.
PREVIEW: LESSON
PERFORMANCE TASK
View the Engage section online. Discuss the photo
and how the number of teams and the attendance are
both functions of time. Then preview the Lesson
Performance Task.
Add.
© Houghton Mifflin Harcourt Publishing Company
Possible answer: You can divide polynomials using
long division or, for a divisor of the form x - a,
synthetic division. The divisor is a factor of the
dividend when the remainder is 0.
12 · -2 = -24
-24 + 1 = -23
Given p(x) = 4x 3 + 3x 2 + 2x + 1, find p(-2) by using synthetic substitution. The dashed
arrow indicates bringing down, the diagonal arrows represent multiplication by –2, and the
solid down arrows indicate adding.
The first two steps are to bring down the leading number, 4, then multiply by the value you
are evaluating at, -2.
-2
4
3
2
1
2
1
-8
4
B
Add 3 and –8.
-2
4
3
-8
-5
4
Module 6
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Lesson 5
365
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HARDCOVER PAGES 281290
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Lesson 5
365
Module 6
6L5 365
85_U3M0
ESE3898
IN3_MNL
365
Lesson 6.5
Resource
Locker
19/07/14
2:20 AM
19/07/14 2:16 AM

Multiply the previous answer by –2.
-2
4
3
-8
Continue this sequence of steps until you reach
the last addition.
1
-2
10
-5
4

2

4
4
3
2
1
-8
10
-24
-5
12
-23
Evaluating a Polynomial Function
Using Synthetic Substitution
p(-2) = -23
INTEGRATE TECHNOLOGY
HARDBOUND SE
Reflect
1.
EXPLORE
PAGE 282
BEGINS HERE
Discussion After the final addition, what does this sum correspond to?
The final sum represents the value of p(x) where x = -2.
QUESTIONING STRATEGIES
Dividing Polynomials Using Long Division
Explain 1
What operation are you doing and what are
you finding when you use synthetic
substitution on the polynomial function p(x) to find
p(a)? You are dividing the polynomial function p(x)
by the quantity (x - a), and you are finding the
value of p(a).
Recall that arithmetic long division proceeds as follows.
Divisor
––––
23 ← Quotient
12 ⟌ 277 ← Dividend
24
―
37
36
―
1 ← Remainder
dividend
remainder
Notice that the long division leads to the result _______
= quotient + ________
. Using the
divisor
divisor
277
1
numbers from above, the arithmetic long division leads to ___
= 23 + __
. Multiplying through
12
12
by the divisor yields the result dividend = (divisor)(quotient) + remainder. (This can be used as

EXPLAIN 1
© Houghton Mifflin Harcourt Publishing Company
a means of checking your work.)
Example 1
Given a polynomial divisor and dividend, use long division to
find the quotient and remainder. Write the result in the form
dividend = (divisor)(quotient) + remainder and then carry out the
multiplication and addition as a check.
(4x 3 + 2x 2 + 3x + 5) ÷ (x 2 + 3x + 1)
Begin by writing the dividend in standard form, including terms with a coefficient
of 0 (if any).
4x 3 + 2x 2 + 3x + 5
Dividing Polynomials Using Long
Division
AVOID COMMON ERRORS
Write division in the same way as you would when dividing numbers.
–––––––––––––––––
x 2 + 3x + 1 ⟌ 4x 3 + 2x 2 + 3x + 5
Module 6
Students have the option of completing the
polynomial division activity either in the book
or online.
366
Lesson 5
PROFESSIONAL DEVELOPMENT
IN3_MNLESE389885_U3M06L5 366
Math Background
Division of polynomials is related to division of whole numbers. Given
P(x)
R(x)
polynomials P(x) and D(x), where D(x) ≠ 0, we can write _____ = Q(x) + _____ ,
D(x)
D(x)
where the remainder R(x) is a polynomial whose degree is less than that of D(x) .
(If the degree of R(x) were not less than the degree of D(x), we would be able to
continue dividing.) Equivalently, P(x) = Q(x)D(x) + R(x). This last expression
can be used to justify the Remainder Theorem. Notice that when D(x) is a linear
divisor of the form x - a, the expression becomes P(x) = Q(x)(x - a) + r, where
the remainder r is a real number.
23/07/14 12:19 AM
Students may have difficulty relating the familiar
long-division process for whole numbers to
identifying the process for polynomials using the
algorithm for finding dividend = (divisor)(quotient)
+ remainder. Point out that polynomial long division
remainder ,
dividend = quotient + _________
leads to this result: ________
divisor
divisor
which is equivalent to dividend = (divisor)(quotient)
+ remainder if you multiply each term by the divisor.
Showing an example of arithmetic long division
alongside an example of polynomial division may
help students make the connection.
Dividing Polynomials 366
Find the value you need to multiply the divisor by so that the first term matches
with the first term of the dividend. In this case, in order to get 4x 2, we must multiply
x 2 by 4x. This will be the first term of the quotient.
4x
QUESTIONING STRATEGIES
–––––––––––––––––
How can you tell if you are finished solving a
polynomial division problem? The remainder
has a degree less than the degree of the divisor, or
has degree 0.
What do you write as the final answer for a
polynomial division problem? The answer
should be written as the product of factors plus the
remainder, where one factor is the divisor and the
other factor is the quotient.
x 2 + 3x + 1 ⟌ 4x 3 + 2x 2 + 3x + 5
Next, multiply the divisor through by the term of the quotient you just found and
subtract that value from the dividend. (x 2 + 3x + 1)(4x) = 4x 3 + 12x 2 + 4x, so
subtract 4x 3 + 12x 2 + 4x from 4x 3 + 2x 2 + 3x + 5.
–––––––––––––––––
4x
x 2 + 3x + 1 ⟌ 4x 3 + 2x 2 + 3x + 5
-(4x 3 + 12x 2 + 4x)
―――――――
-10x 2 - x + 5
HARDBOUND SE
Taking this difference as the new dividend, continue in this fashion until the largest
term of the remaining dividend is of lower degree than the divisor.
PAGE 283
BEGINS HERE
4x - 10
–––––––––––––––––
x 2 + 3x + 1 ⟌ 4x 3 + 2x 2 + 3x + 5
-(4x 3 + 12x 2 + 4x)
―――――――
-10x - x + 5
2
-(-10x 2 - 30x - 10)
――――――――
29x + 15
Since 29x + 5 is of lower degree than x 2 + 3x + 1, stop. 29x + 15 is the remainder.
Write the final answer.
4x 3 + 2x 2 + 3x + 5 = (x 2 + 3x + 1)(4x - 10) + 29x + 15
Check.
4x 3 + 2x 2 + 3x + 5 = (x 2 + 3x + 1)(4x - 10) + 29x + 15
© Houghton Mifflin Harcourt Publishing Company
= 4x 3 + 12x 2 + 4x - 10x 2 - 30x - 10 + 29x + 15
= 4x 3 + 2x 2 + 3x + 5
B
(6x 4 + 5x 3 + 2x + 8) ÷ (x 2 + 2x - 5)
Write the dividend in standard form, including terms with a coefficient of 0.
6x 4 + 5x 3 + 0x 2 + 2x + 8
Write the division in the same way as you would when dividing numbers.
––––––––––––––––––––––
x 2 + 2x - 5 ⟌ 6x 4 + 5x 3 + 0x 2 + 2x + 8
Module 6
367
Lesson 5
COLLABORATIVE LEARNING
IN3_MNLESE389885_U3M06L5 367
Small Group Activity
Help groups of students practice dividing polynomials using synthetic division.
Provide each student with a different example of polynomial division, and ask
them to show and explain the first step in dividing with synthetic division. Then
have them pass the problem to another student, who writes the next step and
explains it. They continue to pass the problem until each problem is completely
solved and all steps are explained. The last student summarizes by giving the
quotient and remainder in polynomial form. Encourage students to use these as
examples of dividing polynomials when they write in their journals.
367
Lesson 6.5
19/07/14 2:44 AM
Divide.
6x 2 - 7x + 44
––––––––––––––––––––––
x 2 + 2x - 5 ⟌ 6x 4 + 5x 3 + 0x 2 + 2x + 8
-(6x 4 + 12x 3 - 30x 2)
――――――――――
(
-7x 3 + 30x 2 + 2x
2
- -7x 3 -14x + 35x
)
―――
44x 2 - 33x + 8
-
( 44x + 88x - 220 )
2
―――
-121x + 228
Write the final answer.
6x 4 + 5x 3 + 2x + 8 = (x 2 + 2x - 5)(6x 2 - 7x + 44) - 121x + 228
HARDBOUND SE
Check.
PAGE 284
6x 4 + 5x 3 + 2x + 8 = (x 2 + 2x - 5)(6x 2 - 7x + 44) -121x + 228
BEGINS HERE
= 6x 4 - 7x 3 + 44x 2 + 12x 3 -14x 2 + 88x - 30x 2 + 35x - 220 - 121x + 228
= 6x 4 + 5x 3 + 2x + 8
Reflect
2.
How do you include the terms with coefficients of 0?
You represent the term with 0 as the coefficient, e.g, 0x.
3.
(15x 3 + 8x - 12) ÷ (3x 2 + 6x + 1)
(3x 2 + 6x + 1)(5x - 10) + 63x - 2
5x - 10
––––––––––––––––––––
3x 2 + 6x + 1 ⟌ 15x 3 + 0x 2 + 8x - 12
= 15x 3 - 30x 2 + 30x 2 - 60x + 5x - 10 +
-(15x 3 + 30x 2 + 5x)
63x - 2
――――――――
= 15x 3 + 8x - 12
-30x + 3x - 12
2
-(-30x 2 - 60x - 10)
© Houghton Mifflin Harcourt Publishing Company
Your Turn
Use long division to find the quotient and remainder. Write the result in the form
dividend = (divisor)(quotient) + remainder and then carry out a check.
―――――――――
63x - 2
Module 6
368
Lesson 5
DIFFERENTIATE INSTRUCTION
IN3_MNLESE389885_U3M06L5 368
Graphic Organizers
19/07/14 3:04 AM
Have groups of students create
graphic organizers to help them
divide polynomials using synthetic
division. Have them show how to
organize a problem into a form
similar to the one shown. Then have
them use organizers to write each of
the steps, explain what goes into each of the cells, and then interpret the results.
Dividing Polynomials 368
(9x 4 + x 3 + 11x 2 - 4) ÷ (x 2 + 16)
4.
EXPLAIN 2
x2
Dividing p(x) by x - a Using
Synthetic Division
9
-4
-6
7+(-4)
2
-2(2)
3
-2(3)
3
)
16x + 2124
2
= 9x 4 + x 3 + 11x 2 - 4
x - 133x + 0x
3
2
-(x 3 + 0x 2 + 16x)
―――――――
-133x 2 - 16x - 4
-(-133x 2 + 0x - 2128)
――――――――――
-16x + 2124
Explain 2
Dividing p(x) by x - a Using Synthetic Division
Compare long division with synthetic substitution. There are two important things to notice. The first
is that p(a) is equal to the remainder when p(x) is divided by x - a. The second is that the numbers to
the left of p(a) in the bottom row of the synthetic substitution array give the coefficients of the quotient.
For this reason, synthetic substitution is also called synthetic division.
Long Division
––––––––––––––––––
Synthetic Substitution
3x 2 + 10x + 20
x - 2 ⟌ 3x 3 + 4x 2 + 0x + 10
-(3x 3 - 6x 2)
10x 2 + 0x
-(10x 2 - 20x)
20x + 10
-20x - 40
――――
50
―――――
2
3 4 0 10
6 20 40
――――――
3 10 20 50
――――――
9+(-6)
3
Students should notice that the divisor is -2 because
the form of the divisor is (x - a), or (x - (-2)); the
rows are added; the remainder is the last digit in the
last row, 3; and the last row includes the coefficients
of the quotient, starting with the power of the
variable decreased by 1.
3
(2x 2 + 7x + 9) ÷ (x + 2) = 2x + 3 + _____
x+2
= 9x 4 + x 3 - 133x 2 + 144x 2 + 16x - 2128 -
2
――――――――
Example 2
HARDBOUND SE
© Houghton Mifflin Harcourt Publishing Company
bring
down 2
7
3
-(9x + 0x + 144x
patterns in synthetic division. Have students use
arrows and expressions, if necessary, to help them
understand the patterns. For example, (2x 2 + 7x + 9)
÷ (x + 2) may be shown as:
2
4
4
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Patterns
MP.8 Students should quickly see that there are
-2
(x 2 + 16)(9x 2 + x - 133) - 16x + 2124
9x 2 + x - 133
+ 16 ⟌––––––––––––––––––––––––
9x + x + 11x + 0x - 4

Given a polynomial p(x), use synthetic division to divide by x - a and obtain
the quotient and the (nonzero) remainder. Write the result in the form
p(x) = (x - a) (quotient) + p(a) then carry out the multiplication and addition
as a check.
(7x 3 - 6x + 9) ÷ (x + 5)
By inspection, a = -5. Write the coefficients
and a in the synthetic division format.
-5
Bring down the first coefficient. Then multiply
and add for each column.
-5
Write the result, using the non-remainder
entries of the bottom row as the coefficients.
PAGE 285
BEGINS HERE
Check.
7 0 -6 9
――――――
0
-6
9
-35 175 -845
―――――――――
7 -35 169 -836
7
(7x 3 - 6x + 9) = (x + 5)(7x 2 - 35x + 169) - 836
(7x 3 - 6x + 9) = (x + 5)(7x 2 - 35x + 169) - 836
= 7x 3 - 35x 2 - 35x 2 - 175x + 169x + 845 - 836
= 7x 3 - 6x + 9
Module 6
369
Lesson 5
LANGUAGE SUPPORT
IN3_MNLESE389885_U3M06L5 369
Connect Vocabulary
Help students understand how the method synthetic division is used as a symbolic
representation of a polynomial division problem. Point out that the English word
synthetic means not genuine, unnatural, artificial, or contrived. That implies they
will have to understand how to interpret the numerical results in the last row of
the synthetic division problem. To help students remember the value of a for the
divisor (x - a), show them how to form the equation x - a = 0, and then solve
that equation for a. This procedure will remind students to use the correct sign for
the divisor.
369
Lesson 6.5
23/07/14 12:27 AM
B
(4x 4 - 3x 2 + 7x + 2) ÷
(x - _21 )
QUESTIONING STRATEGIES
Find a. Then write the coefficients and a in the synthetic division format.
How can you recognize the quotient and
remainder when using the synthetic division
method? The bottom row of the synthetic division
problem gives the coefficients of the quotient along
with the remainder of the division problem.
1
Find a = _
2
1
_
2 4 0 -3 7 2
―――――――
Bring down the first coefficient. Then multiply and add for each column.
1
_
2 4 0 -3 7 2
2 1 -1 3
―――――――
4 2 -2 6 5
Write the result.
(x - _12 )(4x + 2x - 2x + 6) + 5
(4x 4 - 3x 2 + 7x + 2)=
2
3
Check.
(4x 4 - 3x 2 + 7x + 2)=
(x - _12 )(4x + 2x - 2x + 6) + 5
3
2
= 4x 4 + 2x 3 - 2x 2 + 6x - 2x 3 - x 2 + x - 3 + 5
= 4x 4 + 3x 2 + 7x + 2
Reflect
5.
Can you use synthetic division to divide a polynomial by x 2 + 3? Explain.
No, the divisor must be a linear binomial in the form x - a; x 2 + 3 is a quadratic binomial.
Your Turn
6.
(2x 3 + 5x 2 - x + 7) ÷ (x - 2)
2
2
5 -1
4
(x - 2)(2x 2 + 9x + 17) + 41
7
= 2x 3 - 4x 2 + 9x 2 - 18x + 17x - 34 + 41
18 34
――――――
2 9 17 41
7.
(6x 4 - 25x 3 - 3x + 5) ÷
1
-_
3
6 -25 0 -3
= 2x 3 + 5x 2 - x + 7
(x + _31 )
(x + _13 )(6x - 27x + 9x - 6) + 7
3
5
2
© Houghton Mifflin Harcourt Publishing Company
Given a polynomial p(x), use synthetic division to divide by x - a and obtain
the quotient and the (nonzero) remainder. Write the result in the form
p(x) = (x - a)(quotient) + p(a). You may wish to perform a check.
= 6x + 2x - 27x - 9x 2 + 9x 2 + 3x - 6x
-2 9 -3 2
―――――――
6 -27 9 -6 7
4
3
3
-2+7
= 6x 4 - 25x 3 - 3x + 5
Module 6
IN3_MNLESE389885_U3M06L5 370
370
Lesson 5
19/07/14 3:07 AM
Dividing Polynomials 370
HARDBOUND SE
EXPLAIN 3
Using the Remainder Theorem and Factor
Theorem
Explain 3
PAGE 286
BEGINS HERE
When p(x) is divided by x - a, the result can be written in the form p(x) = (x - a)q(x) + r
where q(x) is the quotient and r is a number. Substituting a for x in this equation gives
p(a) = (a - a)q(a) + r. Since a - a = 0, this simplifies to p(a) = r. This is known as
the Remainder Theorem.
Using the Remainder Theorem and
Factor Theorem
If the remainder p(a) in p(x) = (x - a)q(x) + p(a) is 0, then p(x) = (x - a)q(x), which tells you
that x - a is a factor of p(x). Conversely, if x - a is a factor of p(x), then you can write p(x) as
p(x) = (x - a)q(x), and when you divide p(x) by x - a, you get the quotient q(x) with a remainder
of 0. These facts are known as the Factor Theorem.
QUESTIONING STRATEGIES
How are the Remainder Theorem and the
Factor Theorem related? The Remainder
Theorem implies that if a polynomial p(x) is divided
by x - a, and the remainder p(a) = 0, then (x - a) is
a factor of the polynomial. The Factor Theorem says
that if (x - a) is a factor of p(x), you can rewrite the
polynomial as the quotient q(x) times this factor, or
p(x) = (x - a)q(x).
Determine whether the given binomial is a factor of the polynomial p(x). If
so, find the remaining factors of p(x).
Example 3

p(x) = x 3 + 3x 2 - 4x - 12; (x + 3)
Use synthetic division.
-3
3 -4 -12
-3
0
12
――――――――
1 0 -4
0
1
Since the remainder is 0, x + 3 is a factor.
Write q(x) and then factor it.
q(x) = x 2 - 4 = (x + 2)(x - 2)
So, p(x) = x 3 + 3x 2 - 4x - 12 = (x + 2)(x - 2)(x + 3).

p(x) = x 4 - 4x 3 - 6x 2 + 4x + 5; (x + 1)
Use synthetic division.
© Houghton Mifflin Harcourt Publishing Company
-1
-4 -6 4 5
-1
5 1-5
―――――――
-5
-1
5 0
1
1
Since the remainder is
0 , (x + 1) is a factor. Write q(x).
q(x) = x - 5x - x +5
3
2
Now factor q(x) by grouping.
q(x) = x 3 - 5x 2 - x +5
= x 2(x - 5) - (x - 5)
=
(x 2 - 1)(x - 5)
=
(x + 1)(x - 1)(x - 5)
So, p(x) = x 4 - 4x 3 - 6x 2 + 4x + 5 =
Module 6
IN3_MNLESE389885_U3M06L5 371
371
Lesson 6.5
(x + 1)(x + 1)(x - 1)(x - 5)
371
.
Lesson 5
19/07/14 3:08 AM
HARDBOUND SE
Your Turn
PAGE 287
Determine whether the given binomial is a factor of the polynomial p(x). If it is, find
the remaining factors of p(x).
8.
p(x) = 2x 4 + 8x 3 + 2x + 8; (x + 4)
-4
9.
2
8 0 2 8
-8 0 0 -8
――――――――
2 0 0 2
0
3 0 -2 5
3
3 1
――――――
―――――
3 3
1 6
ELABORATE
QUESTIONING STRATEGIES
Since the remainder is 0, (x + 4) is a factor.
When do you use synthetic substitution, and
when do you use synthetic division? You use
synthetic substitution when you want to find the
function value of a polynomial for a certain number.
You use synthetic division when you want to find
the quotient of polynomial division.
q(x) = 2x 3 + 2 = 2(x 3 + 1) = 2(x + 1)(x 2 - x + 1)
So, p(x) = 2x 4 + 8x 3 + 2x + 8
= 2(x + 1)(x 2 - x + 1)(x + 4)
p(x) = 3x 3 - 2x + 5; (x - 1)
1
BEGINS HERE
Since the remainder is 6, x - 1 is not a factor.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Communication
MP.3 Have students work in pairs to complete a
Elaborate
10. Compare long division and synthetic division of polynomials.
The numbers generated by synthetic division are equal to the coefficients of the terms of
the polynomial quotient, including the remainder. They are essentially the same process.
chart like the following, showing similarities and
differences:
11. How does knowing one linear factor of a polynomial help find the other factors?
If one linear factor of the polynomial is known, synthetic division can be used to find the
Alike
Long Division
product of the other factors, which may be easily factorable.
written in standard form with 0 representing any missing terms.
13. Essential Question Check-In How do you know when the divisor is a factor of the dividend?
The divisor is a factor of the dividend when the remainder is 0.
IN3_MNLESE389885_U3M06L5 372
372
Synthetic
Substitution
or Division
© Houghton Mifflin Harcourt Publishing Company
12. What conditions must be met in order to use synthetic division?
The divisor must be a linear binomial with a leading coefficient of 1. The dividend must be
Module 6
Different
SUMMARIZE THE LESSON
How do you explain the process of synthetic
division, and when and why it is useful?
What are possible sources of error in the
process? Synthetic division is a division process that
uses only the coefficients of a polynomial. If the last
sum is 0, then the binomial is a factor of the
polynomial; possible errors are not bringing down
the first coefficient, forgetting to add instead of
subtract, and forgetting to include coefficients
that are 0.
Lesson 5
19/07/14 3:09 AM
Dividing Polynomials 372
Evaluate: Homework and Practice
EVALUATE
• Online Homework
• Hints and Help
• Extra Practice
Given p(x), find p(-3) by using synthetic substitution.
p(x) = 8x 3 + 7x 2 + 2x + 4
1.
-3
8
Concepts and Skills
Practice
Explore
Evaluating a Polynomial Function
Using Synthetic Substitution
Exercises 1–4
Example 1
Dividing Polynomials Using Long
Division
Exercises 5–8
Example 2
Dividing p(x) by x - a Using
Synthetic Division
Exercises 9–11
Example 3
Using the Remainder Theorem and
Factor Theorem
Exercises 12–15
2
-3
4
-24
51 -159
-17
53 -155
p(-3) = -155
p(x) = 2x 3 + 5x 2 - 3x
3.
-3
2
p(-3) = 0
Students might make errors in signs when doing
synthetic division and synthetic substitution because
values are added rather than subtracted as in long
division. Remind them that terms are always added
for synthetic substitution and synthetic division.
5.
3
0
-1
0
0
-3
-9
-25
1
3
6
-2 -19
p(-3) = -19
p(x) = -x 4 + 5x 3 - 8x + 45
-3
-1
5
0
-8
45
3 -24
72 -192
――――――――――――
-1
8 -24
64
-147
p(-3) = -147
(18x 3 - 3x 2 + x -1) ÷ (x 2 - 4)
18x - 3
2
Check.
- 4 ⟌–––––––––––––––––––
18x - 3x + x - 1
3
(x 2 - 4)(18x - 3) + 73x - 13
2
-(18x 3 + 0x 2 - 72x)
= 18x 3 - 72x - 3x 2 + 12 + 73x - 13
―――――――――
-3x 2 + 73x - 1
= 18x 3 - 3x 2 + x - 1
-(-3x 2 + 0x + 12)
――――――――
73x - 13
6.
(6x 4 + x 3 - 9x + 13) ÷ (x 2 + 8)
6x 2 + x - 48
x
2
+ 8 ⟌–––––––––––––––––––––––––
6x + x + 0x - 9x + 13
4
3
2
-(6x 4 + 0x 3 + 48x 2)
――――――――――――
x 3 - 48x 2 - 9x
AVOID COMMON ERRORS
Students may be confused about when to use
synthetic division and when to use long division.
Point out that for a divisor other than a linear
binomial with leading coefficient 1, long division is
the best method. It may, however, be possible to
divide the dividend and divisor by a constant to make
the leading coefficient 1.
-6
0
7
Given a polynomial divisor and dividend, use long division to find the quotient and
remainder. Write the result in the form dividend = (divisor)(quotient) + remainder.
You may wish to carry out a check.
© Houghton Mifflin Harcourt Publishing Company
AVOID COMMON ERRORS
-3
6
―――――――
4.
5
2
1
――――――――
x
Lesson 6.5
7
p(x) = x 3 + 6x 2 + 7x - 25
――――――――
ASSIGNMENT GUIDE
373
8
2.
-(x + 0x + 8x)
Check.
(x 2 + 8)(6x 2 + x - 48) - 17x + 397
= 6x 4 + x 3 - 48x 2 + 48x 2 + 8x - 384 - 17x + 397
= 6x 4 + x 3 - 9x + 13
―――――――――
3
2
-48x 2 - 17x + 13
-(-48x 2 + 0x - 384)
―――――――――
-17x + 397
Module 6
Exercise
IN3_MNLESE389885_U3M06L5.indd 373
Lesson 5
373
Depth of Knowledge (D.O.K.)
Mathematical Practices
1–15
1 Recall of Information
MP.2 Reasoning
16–19
2 Skills/Concepts
MP.4 Modeling
20
2 Skills/Concepts
MP.2 Reasoning
21
3 Strategic Thinking
MP.2 Reasoning
22
3 Strategic Thinking
MP.6 Precision
23
3 Strategic Thinking
MP.2 Reasoning
4/7/14 3:18 PM
7.
(x 4 + 6x - 2.5) ÷ (x 2 + 3x + 0.5)
x 2 - 3x + 8.5
––––––––––––––––––––––––
( x 2 + 3x + 0.5)(x 2 - 3x + 8.5) - 18x - 6.75
x 2 + 3x + 0.5 ⟌ x 4 + 0x 3 + 0x 2 + 6x - 2.5
-(x 4 + 3x 3 + 0.5x 2)
A common error when doing synthetic division is to
subtract the second row rather than adding it. Show
students a long division problem alongside its
solution, using synthetic division to emphasize that
the results will be different if they make this error.
= x 4 - 3x 3 + 8.5x 2 + 3x 3 - 9x 2 + 25.5x + 0.5x 2
―――――――――――
-3x 3 - 0.5x 2 + 6x
- 1.5x + 4.25 - 18x - 6.75
-(-3x - 9x - 1.5x)
= x 4 + 6x - 2.5
―――――――――
3
AVOID COMMON ERRORS
Check.
2
8.5x 2 + 7.5x - 2.5
-(8.5x 2 + 25.5x + 4.25)
―――――――――
8.
-18x - 6.75
1
2
_
(x + 250x + 100x) ÷ 2 x + 25x + 9
2x + 400
1 2
x + 25x + 9 ⟌ x 3 + 250x 2 + 100x + 0
2
-(x 3 + 50x 2 + 18x)
3
(
2
)
Check.
1 2
x + 25x + 9 (2x + 400) - 9918x - 3600
2
3
= x + 200x 2 + 50x 2 + 10, 000x + 18x + 3600
–––––––––––––––––––––
_
(_
――――――――
200x 2 + 82x + 0
-(200x 2 + 10, 000x + 3600)
―――――――――――
)
- 9918x - 3600
= x 3 + 250x 2 + 100x
-9918x - 3600
Given a polynomial p(x), use synthetic division to divide by x - a and obtain
the quotient and the (nonzero) remainder. Write the result in the form
p(x) = (x - a)(quotient) + p(a). You may wish to carry out a check.
9.
(7x 3 - 4x 2 - 400x - 100) ÷ (x - 8)
(x - 8)(7x 2 + 52x + 16) + 28
-4 -400 -100
56
416
128
―――――――――
7 52
16
28
8
7
= 7x 3 - 56x 2 + 52x 2 - 416x + 16x - 128 + 28
= 7x 3 - 4x 2 - 400x - 100
-0.25
8
0
-28.5
-9
10
-2
0.5
7
-0.5
―――――――――――――
8 -2
-28
-2
9.5
11.
(2.5x 3 + 6x 2 - 5.5x - 10) ÷ (x + 1)
-1
IN3_MNLESE389885_U3M06L5 374
= 8x 4 + 2x 3 - 2x 3 - 0.5x 2 - 28x 2 - 7x - 2x
- 0.5 + 9.5
= 8x 4 - 28.5x 2 - 9x + 10
(x + 1)(2.5x 2 + 3.5x - 9) - 1
6 -5.5 -10
-2.5 -3.5
9
―――――――――
2.5 3.5
-9
-1
2.5
Module 6
(x + 0.25) (8x 3 - 2x 2 - 28x - 2) + 9.5
© Houghton Mifflin Harcourt Publishing Company
10. (8x 4 - 28.5x 2 - 9x + 10) ÷ (x + 0.25)
= 2.5x 3 + 2.5x 2 + 3.5x 2 + 3.5x - 9x - 9 - 1
= 2.5x 3 + 6x 2 - 5.5x - 10
374
Lesson 5
8/7/14 7:25 AM
Dividing Polynomials 374
HARDBOUND SE
Determine whether the given binomial is a factor of the polynomial p(x) .
If so, find the remaining factors of p(x).
PAGE 288
BEGINS HERE
12. p(x) = x 3 + 2x 2 - x - 2; (x + 2)
-2
13. p(x) = 2x 4 + 6x 3 - 5x - 10; (x + 2)
-2
2 -1 -2
-2
0
2
――――――
1 0 -1
0
1
x + 2 is a factor.
6
0 -5
-10
-4 -4
8
-6
――――――――――
2
2 -4
3 -16
2
x + 2 is not a factor.
x - 1 = (x + 1)(x - 1)
2
So, p(x) = x 3 + 2x 2 - x - 2
= (x + 1)(x - 1)(x + 2).
14. p(x) = x - 22x 2 + 157x - 360; (x - 8)
3
8
1 -22 157 -360
8 -112
360
――――――――――
1 -14
45
0 x - 8 is a factor.
x 2 - 14x + 45 = (x - 5)(x - 9)
So, p(x) = x 3 - 22x 2 + 157x - 360 = (x - 5)(x - 9)(x - 8).
15. p(x) = 4x 3 - 12x 2 + 2x - 5; (x - 3)
© Houghton Mifflin Harcourt Publishing Company
3
4 -12
2 -5
12 0
6
――――――――
4
0 2
1 x - 3 is not a factor.
16. The volume of a rectangular prism whose dimensions are binomials with integer
coefficients is modeled by the function V(x) = x 3 - 8x 2 + 19x - 12 . Given that
x - 1 and x - 3 are two of the dimensions, find the missing dimension of
the prism.
1
1 -8
19 -12
1 -7
12
――――――――
1 -7
12
0
This gives the expression x 2 -7x + 12.
3
1 -7
12
3 -12
―――――
1 -4
0
This gives the expression x - 4, which is the missing dimension.
Module 6
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375
Lesson 6.5
375
Lesson 5
3/11/16 4:48 AM
17. Given that the height of a rectangular prism is x + 2 and the volume is x 3 - x 2 - 6x,
write an expression that represents the area of the base of the prism.
-2
1
-1
-6
0
-2
6
0
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Emphasize the conditions that must be met
――――――――
1 -3
0
0
to use synthetic division: The divisor must be a linear
binomial with a leading coefficient of 1. The dividend
must be written in standard form with 0 representing
any missing terms.
So, the area can be represented by x 2 - 3x.
18. Physics A Van de Graaff generator is a machine
that produces very high voltages by using small, safe
levels of electric current. One machine has a current
that can be modeled by l(t) = t + 2 , where t > 0
represents time in seconds. The power of the system
can be modeled by P(t) = 0.5t 3 + 6t 2 + 10t. Write an
expression that represents the voltage of the system.
Recall that V = __Pl .
-2
Discuss the characteristics of synthetic division that
make it synthetic: there are no variables, only
coefficients; and addition is used instead of
subtraction.
0.5 6 10 0
-1 -10 0
―――――――
0.5 5
0 0
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Reasoning
MP.2 Point out that students should review how to
The voltage can be represented by 0.5t 2 + 5t.
19. Geometry The volume of a hexagonal pyramid is modeled by the function
2
V(x) = __13 x 3 + __43 x + __23 x - __13 . Given the height x + 1, use polynomial division to find
an expression for the area of the base.
© Houghton Mifflin Harcourt Publishing Company • ©Ted Kinsman/Science
Photo Library
(Hint: For a pyramid, V = __13 Bh.)
1 3
(x + 4x 2 + 2x - 1).
V(x) = _
3
-1
1 4 2 -1
-1 -3
1
――――――
1 3 -1
0
So, the area of the base can be represented by x 2 + 3x - 1 .
20. Explain the Error Two students used synthetic division to divide 3x 3 - 2x - 8
by x - 2. Determine which solution is correct. Find the error in the other solution.
A.
2
3 0 -2 -8
6 12 20
―――――――
3 6 10 12
B.
-2
0 -2 -8
-6 12 -20
―――――――
3 -6 10 -28
3
do synthetic division and long division when the
dividend is missing terms for some powers of the
variable. Emphasize that they must include the
missing terms written as zero times the power of the
variable to complete the division.
Student A is correct. Student B used the incorrect sign of a.
Module 6
IN3_MNLESE389885_U3M06L5 376
376
Lesson 5
09/06/15 1:43 PM
Dividing Polynomials 376
PEERTOPEER DISCUSSION
Instruct one student in each pair to solve a
polynomial division problem using long division,
while the other student solves it by using synthetic
division. Then have students switch roles and repeat
the exercise for a new division problem with a
polynomial that does not have a linear factor. Then
have them discuss their results and any preferences
they have for one method or the other.
HARDBOUND SE
H.O.T. Focus on Higher Order Thinking
PAGE 289
BEGINS HERE
21. Multi-Step Use synthetic division to divide p(x) = 3x 3 - 11x 2 - 56x - 50
by (3x + 4). Then check the solution.
(
Rewrite 3x + 4 as 3 x +
4
-_
3
3
-11
_4 ).
3
-56
-50
-4
20
48
―――――――――
3
-15
-36
-2
The quotient needs to be divided by 3.
3x - 15x - 36
__
=x
2
3
Check.
JOURNAL
(
2
- 5x - 12
)
3x 2 - 15x - 36
-2
(3x + 4) __
Have students describe a mnemonic device that can
help them remember the steps in synthetic division.
3
= (3x + 4)(x 2 - 5x - 12) - 2
= 3x 3 + 4x 2 - 15x 2 - 20x - 36x - 48 - 2
= 3x 3 - 11x 2 - 56x - 50
22. Critical Thinking The polynomial ax 3 + bx 2 + cx + d is factored as
3(x - 2)(x + 3)(x - 4). What are the values of a and d? Explain.
© Houghton Mifflin Harcourt Publishing Company
a = 3; d = 72; The value of a is the product of the GCF and the coefficients
of the x-terms of each factor. The value of d is the product of the GCF and
the constant terms of each factor.
23. Analyze Relationships Investigate whether the set of whole numbers, the set
of integers, and the set of rational numbers are closed under each of the four basic
operations. Then consider whether the set of polynomials in one variable is closed
under the four basic operations, and determine whether polynomials are like whole
numbers, integers, or rational numbers with respect to closure. Use the table to
organize.
Whole
Numbers
Integers
Rational
Numbers
Polynomials
Addition
Yes
Yes
Yes
Yes
Subtraction
No
Yes
Yes
Yes
Multiplication
Yes
Yes
Yes
Yes
Division (by
nonzero)
No
No
Yes
No
Polynomials are similar to integers with respect to closure.
They are closed under each operation except division.
Module 6
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377
Lesson 6.5
377
Lesson 5
3/11/16 4:48 AM
Lesson Performance Task
HARDBOUND SE
PAGE 290
BEGINS HERE
The table gives the attendance data for all divisions of NCAA Women’s Basketball.
Students may try to include the “3 divisions” in their
calculation, perhaps by dividing the number of teams
or attendance by 3. Explain to students that a division
is a group of schools that compete against each other
and there are 3 such groups, or divisions, for college
basketball. Explain that this number is irrelevant for
this calculation. What is important is that the
numbers in the table represent the total number of
teams and the total attendance for all of women’s
collegiate basketball.
NCAA Women’s Basketball Attendance
2006–2007
Years since
2006–2007
0
2007–2008
1
Season
Attendance
Number of teams (in thousands) for
in all 3 divisions
all 3 divisions
1003
10,878.3
1013
11,120.8
2008–2009
2
1032
11,160.3
2009–2010
3
1037
11,134.7
2010–2011
4
1048
11,160.0
2011–2012
5
1055
11,201.8
Enter the data from the second, third, and fourth columns of the table and perform linear
regression on the data pairs (t, T) and cubic regression on the data pairs (t, A) where t = years
since the 2006–2007 season, T = number of teams, and A = attendance (in thousands).
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Reasoning
MP.2 Have students highlight the remainder in the
A(t)
. Carry out the division to
Then create a model for the average attendance per team: A avg(t) = ___
T(t)
remainder
write A avg(t) in the form quadratic quotient + _______
.
()
Tt
Use an online computer algebra system to carry out the division of A(t) by T(t).
Models:
final function Aavg(t). Ask them if the remainder is
zero or nonzero and what a nonzero remainder
means. Have students discuss the significance of a
remainder for a function that describes attendance
per team. If the attendance is 315.8 fans per team, ask
students if the 0.8 represents an actual person or if it
results from a limitation of the model.
T(t) = 10.57t + 1005
A(t) = 13.80t 3 - 121.6t 2 + 329.8t + 10,880
Online computer algebra system result:
12,981,600
A avg(t) = 1.30558t 2 - 135.64t + 12,927.9 - __
10.57t + 1005
© Houghton Mifflin Harcourt Publishing Company
Module 6
378
AVOID COMMON ERRORS
Lesson 5
EXTENSION ACTIVITY
IN3_MNLESE389885_U3M06L5.indd 378
Have students research the attendance for a specific NCAA women’s basketball
team for a single season. Have them compare that number to the value calculated
from the model Aavg(t) for the same year. Ask the students how accurate the model
is in predicting the attendance for that team and what some of the sources of error
might be
2/29/16 10:37 PM
Scoring Rubric
2 points: Student correctly solves the problem and explains his/her reasoning.
1 point: Student shows good understanding of the problem but does not fully
solve or explain his/her reasoning.
0 points: Student does not demonstrate understanding of the problem.
Dividing Polynomials 378